WAEC SSCE - Physics - 2001 (Essay)
Question 1 Report
(a) Define specific heat capacity.
(b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using a copper ball.
(ii) State two precautions necessary to obtain accurate results.
(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg\(^{-1}\) K\(^{-1}\)] [specific heat capacity of copper = 400 J kg\(^{-1}\) K\(^{-1}\)]
(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass (1 kg) of a substance by one kelvin (1 K). Its unit is \(\text{J kg}^{-1}\,\text{K}^{-1}\).
(b)(i) Experiment to determine the specific heat capacity of copper (method of mixtures)
Procedure:
- Weigh the empty copper calorimeter together with its stirrer; record the mass \(m_c\).
- Half-fill the calorimeter with water, weigh again, and obtain the mass of water \(m_w\) by subtraction. Note the initial temperature of the water and calorimeter, \(\theta_1\).
- Weigh the copper ball; record its mass \(m_b\). Suspend it by a thread inside a beaker of water and heat until the water boils steadily. Record the steady temperature of the ball, \(\theta\) (about \(100^\circ\text{C}\)).
- Shake off the adhering water and quickly transfer the hot ball into the calorimeter. Stir gently and record the highest steady temperature of the mixture, \(\theta_2\).
Since heat losses are negligible, heat lost by the copper ball = heat gained by the water + heat gained by the calorimeter and stirrer:
\[ m_b c_b(\theta-\theta_2) = m_w c_w(\theta_2-\theta_1) + m_c c_b(\theta_2-\theta_1) \]Making \(c_b\), the specific heat capacity of copper, the subject:
\[ c_b = \frac{m_w c_w(\theta_2-\theta_1)}{m_b(\theta-\theta_2) - m_c(\theta_2-\theta_1)} \](b)(ii) Two precautions
- Transfer the hot copper ball to the calorimeter quickly and shake off any adhering water first, to minimise heat loss to the surroundings.
- Lag (insulate) the calorimeter and stir the water gently before reading the highest steady temperature.
(c) Final steady temperature of the mixture
Data: copper block \(m_b=24\text{ g}=0.024\text{ kg}\) at \(230^\circ\text{C}\); calorimeter \(m_c=60\text{ g}=0.060\text{ kg}\); water \(m_w=54\text{ g}=0.054\text{ kg}\) at \(31^\circ\text{C}\); \(c_w=4200\ \text{J kg}^{-1}\text{K}^{-1}\); \(c_b=400\ \text{J kg}^{-1}\text{K}^{-1}\). Let the final steady temperature be \(\theta\).
Heat lost by copper block = heat gained by water + heat gained by copper calorimeter:
\[ (0.024)(400)(230-\theta) = (0.054)(4200)(\theta-31) + (0.060)(400)(\theta-31) \] \[ 9.6(230-\theta) = (226.8 + 24)(\theta-31) = 250.8(\theta-31) \] \[ 2208 - 9.6\theta = 250.8\theta - 7774.8 \] \[ 2208 + 7774.8 = 250.8\theta + 9.6\theta \] \[ 9982.8 = 260.4\,\theta \] \[ \theta = \frac{9982.8}{260.4} = 38.34^\circ\text{C} \]The final steady temperature of the mixture is \(\theta \approx 38.3^\circ\text{C}\).
Answer Details
(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass (1 kg) of a substance by one kelvin (1 K). Its unit is \(\text{J kg}^{-1}\,\text{K}^{-1}\).
(b)(i) Experiment to determine the specific heat capacity of copper (method of mixtures)
Procedure:
- Weigh the empty copper calorimeter together with its stirrer; record the mass \(m_c\).
- Half-fill the calorimeter with water, weigh again, and obtain the mass of water \(m_w\) by subtraction. Note the initial temperature of the water and calorimeter, \(\theta_1\).
- Weigh the copper ball; record its mass \(m_b\). Suspend it by a thread inside a beaker of water and heat until the water boils steadily. Record the steady temperature of the ball, \(\theta\) (about \(100^\circ\text{C}\)).
- Shake off the adhering water and quickly transfer the hot ball into the calorimeter. Stir gently and record the highest steady temperature of the mixture, \(\theta_2\).
Since heat losses are negligible, heat lost by the copper ball = heat gained by the water + heat gained by the calorimeter and stirrer:
\[ m_b c_b(\theta-\theta_2) = m_w c_w(\theta_2-\theta_1) + m_c c_b(\theta_2-\theta_1) \]Making \(c_b\), the specific heat capacity of copper, the subject:
\[ c_b = \frac{m_w c_w(\theta_2-\theta_1)}{m_b(\theta-\theta_2) - m_c(\theta_2-\theta_1)} \](b)(ii) Two precautions
- Transfer the hot copper ball to the calorimeter quickly and shake off any adhering water first, to minimise heat loss to the surroundings.
- Lag (insulate) the calorimeter and stir the water gently before reading the highest steady temperature.
(c) Final steady temperature of the mixture
Data: copper block \(m_b=24\text{ g}=0.024\text{ kg}\) at \(230^\circ\text{C}\); calorimeter \(m_c=60\text{ g}=0.060\text{ kg}\); water \(m_w=54\text{ g}=0.054\text{ kg}\) at \(31^\circ\text{C}\); \(c_w=4200\ \text{J kg}^{-1}\text{K}^{-1}\); \(c_b=400\ \text{J kg}^{-1}\text{K}^{-1}\). Let the final steady temperature be \(\theta\).
Heat lost by copper block = heat gained by water + heat gained by copper calorimeter:
\[ (0.024)(400)(230-\theta) = (0.054)(4200)(\theta-31) + (0.060)(400)(\theta-31) \] \[ 9.6(230-\theta) = (226.8 + 24)(\theta-31) = 250.8(\theta-31) \] \[ 2208 - 9.6\theta = 250.8\theta - 7774.8 \] \[ 2208 + 7774.8 = 250.8\theta + 9.6\theta \] \[ 9982.8 = 260.4\,\theta \] \[ \theta = \frac{9982.8}{260.4} = 38.34^\circ\text{C} \]The final steady temperature of the mixture is \(\theta \approx 38.3^\circ\text{C}\).
Question 2 Report
The above reprt the graph electron energy against the frequency of the radiation incident on a metal surface. lnterprete the;
(i) slope of the graph;
(ii) intercept, OC;
(iii) intercept, OK.
The graph is a straight line of the maximum kinetic energy \(E\) of the emitted photoelectrons (vertical axis, in joules) against the frequency \(f\) of the incident radiation (horizontal axis, in hertz). The line crosses the frequency axis at the point \(K\) and, when produced backwards, cuts the energy axis below the origin at the point \(C\). The graph is a plot of Einstein's photoelectric equation,
\[ E = hf - hf_0 \]
where \(h\) is Planck's constant and \(f_0\) is the threshold frequency of the metal. Comparing this with the equation of a straight line \(y = mx + c\): \(m = h\) is the gradient and \(c = -hf_0\) is the intercept on the energy axis.
(i) The slope of the graph
Comparing \(E = hf - hf_0\) with \(y = mx + c\), the gradient is
\[ \text{slope} = \frac{\Delta E}{\Delta f} = h \]
So the slope represents Planck's constant, \(h\) \(\left(\approx 6.6\times10^{-34}\,\text{J s}\right)\). It has the same value for every metal.
(ii) The intercept OC
OC is the intercept on the energy axis, obtained when \(f = 0\). From the equation, \(E = -hf_0\), so this intercept is negative and equal in magnitude to the work function \(\varphi\) of the metal:
\[ OC = -hf_0 = -\varphi \]
Thus \(|OC|\) is the minimum energy needed to release an electron from the surface of the metal, i.e. the work function.
(iii) The intercept OK
OK is the intercept on the frequency axis, obtained when the photoelectron energy \(E = 0\). Setting \(0 = hf - hf_0\) gives \(f = f_0\). Therefore OK represents the threshold (cut-off) frequency \(f_0\), the minimum frequency of the incident radiation below which no electrons are emitted from the metal.
Answer Details
The graph is a straight line of the maximum kinetic energy \(E\) of the emitted photoelectrons (vertical axis, in joules) against the frequency \(f\) of the incident radiation (horizontal axis, in hertz). The line crosses the frequency axis at the point \(K\) and, when produced backwards, cuts the energy axis below the origin at the point \(C\). The graph is a plot of Einstein's photoelectric equation,
\[ E = hf - hf_0 \]
where \(h\) is Planck's constant and \(f_0\) is the threshold frequency of the metal. Comparing this with the equation of a straight line \(y = mx + c\): \(m = h\) is the gradient and \(c = -hf_0\) is the intercept on the energy axis.
(i) The slope of the graph
Comparing \(E = hf - hf_0\) with \(y = mx + c\), the gradient is
\[ \text{slope} = \frac{\Delta E}{\Delta f} = h \]
So the slope represents Planck's constant, \(h\) \(\left(\approx 6.6\times10^{-34}\,\text{J s}\right)\). It has the same value for every metal.
(ii) The intercept OC
OC is the intercept on the energy axis, obtained when \(f = 0\). From the equation, \(E = -hf_0\), so this intercept is negative and equal in magnitude to the work function \(\varphi\) of the metal:
\[ OC = -hf_0 = -\varphi \]
Thus \(|OC|\) is the minimum energy needed to release an electron from the surface of the metal, i.e. the work function.
(iii) The intercept OK
OK is the intercept on the frequency axis, obtained when the photoelectron energy \(E = 0\). Setting \(0 = hf - hf_0\) gives \(f = f_0\). Therefore OK represents the threshold (cut-off) frequency \(f_0\), the minimum frequency of the incident radiation below which no electrons are emitted from the metal.
Question 3 Report
A ball thrown vertically upward reaches a maximum height of 50 m above the level of projection. Calculate the;
(i) time taken to reach the maximum height,
(ii) speed of the throw. [g = 10 ms\(^{-2}\)]
Given: maximum height \(h = 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\), and velocity at the top \(= 0\).
(ii) Speed of throw (find first): using \(v^{2} = u^{2} - 2gh\) with \(v = 0\):
\[ 0 = u^{2} - 2(10)(50) \Rightarrow u^{2} = 1000 \] \[ u = \sqrt{1000} = 31.6\ \text{m s}^{-1} \](i) Time to reach maximum height: using \(v = u - gt\) with \(v = 0\):
\[ t = \frac{u}{g} = \frac{31.6}{10} = 3.16\ \text{s} \]The ball takes about 3.16 s to reach the top, having been thrown at about 31.6 m s\(^{-1}\).
Answer Details
Given: maximum height \(h = 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\), and velocity at the top \(= 0\).
(ii) Speed of throw (find first): using \(v^{2} = u^{2} - 2gh\) with \(v = 0\):
\[ 0 = u^{2} - 2(10)(50) \Rightarrow u^{2} = 1000 \] \[ u = \sqrt{1000} = 31.6\ \text{m s}^{-1} \](i) Time to reach maximum height: using \(v = u - gt\) with \(v = 0\):
\[ t = \frac{u}{g} = \frac{31.6}{10} = 3.16\ \text{s} \]The ball takes about 3.16 s to reach the top, having been thrown at about 31.6 m s\(^{-1}\).
Question 4 Report
A lead shot is projected from tha ground level with a velocity u at an angle \(\theta\) to the horizontal. Given the time, t for the lead shot to reach its maximum height as; t = \(\frac{u \sin \theta}{g}\) where "g" is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h\(_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}\)
For the vertical motion of the projectile, the initial vertical velocity is \(u\sin\theta\) and the acceleration due to gravity is \(-g\).
Using
\[ h=(u\sin\theta)t-\frac{1}{2}gt^{2} \]At maximum height,
\[ t=\frac{u\sin\theta}{g} \]Therefore,
\[ h_{\text{max}}=(u\sin\theta)\left(\frac{u\sin\theta}{g}\right) -\frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{u^2\sin^2\theta}{2g} \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{2g} \]Hence, the greatest height reached is
\[ \boxed{h_{\text{max}}=\frac{u^2\sin^2\theta}{2g}} \]Answer Details
For the vertical motion of the projectile, the initial vertical velocity is \(u\sin\theta\) and the acceleration due to gravity is \(-g\).
Using
\[ h=(u\sin\theta)t-\frac{1}{2}gt^{2} \]At maximum height,
\[ t=\frac{u\sin\theta}{g} \]Therefore,
\[ h_{\text{max}}=(u\sin\theta)\left(\frac{u\sin\theta}{g}\right) -\frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{u^2\sin^2\theta}{2g} \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{2g} \]Hence, the greatest height reached is
\[ \boxed{h_{\text{max}}=\frac{u^2\sin^2\theta}{2g}} \]Question 5 Report
(a) Distinguish between stress and strain as used in elasticity.
(b) When a force of 40 N is applied to the free end of an elastic cord, an extension of 5cm is produced in the cord. Calculate the work done on the cord.
(a) Stress and strain
Stress is the force acting per unit cross-sectional area of the material:
\[ \text{stress} = \frac{\text{force}}{\text{area}} \quad (\text{unit: N m}^{-2}\ \text{or Pa}) \]Strain is the extension produced per unit original length of the material:
\[ \text{strain} = \frac{\text{extension}}{\text{original length}} \quad (\text{no unit, it is a ratio}) \]Thus stress measures the intensity of the applied load while strain measures the relative deformation it produces.
(b) Work done on the cord
For an elastic cord obeying Hooke's law, the work done equals the elastic potential energy stored, given by the area under the force-extension graph:
\[ W = \tfrac{1}{2}Fe \]With \(F = 40\,\text{N}\) and \(e = 5\,\text{cm} = 0.05\,\text{m}\):
\[ W = \tfrac{1}{2}(40)(0.05) = 1.0\,\text{J} \]Answer Details
(a) Stress and strain
Stress is the force acting per unit cross-sectional area of the material:
\[ \text{stress} = \frac{\text{force}}{\text{area}} \quad (\text{unit: N m}^{-2}\ \text{or Pa}) \]Strain is the extension produced per unit original length of the material:
\[ \text{strain} = \frac{\text{extension}}{\text{original length}} \quad (\text{no unit, it is a ratio}) \]Thus stress measures the intensity of the applied load while strain measures the relative deformation it produces.
(b) Work done on the cord
For an elastic cord obeying Hooke's law, the work done equals the elastic potential energy stored, given by the area under the force-extension graph:
\[ W = \tfrac{1}{2}Fe \]With \(F = 40\,\text{N}\) and \(e = 5\,\text{cm} = 0.05\,\text{m}\):
\[ W = \tfrac{1}{2}(40)(0.05) = 1.0\,\text{J} \]Question 6 Report
(a) Explain the statement the capacitance of a capacitor is 5\(\mu\)F.
(b)(i) State the factors upon which the capacitance of a parallel plate capacitor depend.
(ii) State how the capacitance depends on each of these factors stated in (b)(i).
(c) A series arrangement of three capacitors of values 8uF, 12\(\mu\)F, and 24\(\mu\)F is connected in series with 90-V battery.
(i) Draw an open-circuit diagram for this arrangement.
(ii) Calculate the effective capacitance in the circuit.
(iii) On closed circuit, calculate the charge on each capacitor when fully charged.
(iv) Determine the p.d across the 8\(\mu\)F capacitor.
(a) To say that the capacitance of a capacitor is \(5\,\mu\text{F}\) means that the capacitor stores a charge of \(5\,\mu\text{C}\) \((5\times10^{-6}\,\text{C})\) on each plate for every \(1\,\text{V}\) of potential difference applied across its plates. In other words, the ratio of the charge stored to the potential difference across the plates is \(5\,\mu\text{C V}^{-1}\), since \(C=\dfrac{Q}{V}\).
(b)(i) & (ii) The capacitance of a parallel-plate capacitor is given by \(C=\dfrac{\varepsilon A}{d}\). The factors on which it depends and the form of dependence are:
| Factor | How the capacitance depends on it |
| Common (overlapping) area of the plates, \(A\) | Directly proportional: \(C\propto A\). Increasing the area increases the capacitance. |
| Distance between the plates, \(d\) | Inversely proportional: \(C\propto \dfrac{1}{d}\). Increasing the separation decreases the capacitance. |
| Permittivity (nature) of the dielectric, \(\varepsilon\) | Directly proportional: \(C\propto \varepsilon\). A dielectric of higher permittivity increases the capacitance. |
(c)(i) Open-circuit diagram (switch open) showing the three capacitors in series with the \(90\,\text{V}\) battery:
(c)(ii) Effective capacitance. For capacitors in series the reciprocals add:
\[\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{3+2+1}{24}=\frac{6}{24}=\frac{1}{4}\]\[C=4\,\mu\text{F}\](c)(iii) Charge on each capacitor. In a series arrangement the same charge flows onto every capacitor, and it equals the charge on the equivalent capacitor:
\[Q=CV=4\times10^{-6}\times90=3.6\times10^{-4}\,\text{C}=360\,\mu\text{C}\]Therefore the charge on each of the \(8\,\mu\text{F}\), \(12\,\mu\text{F}\) and \(24\,\mu\text{F}\) capacitors is the same: \(Q=360\,\mu\text{C}\).
(c)(iv) P.d. across the \(8\,\mu\text{F}\) capacitor.
\[V_{8}=\frac{Q}{C_1}=\frac{360\,\mu\text{C}}{8\,\mu\text{F}}=45\,\text{V}\]The potential difference across the \(8\,\mu\text{F}\) capacitor is \(45\,\text{V}\). (As a check: \(V_{12}=360/12=30\,\text{V}\) and \(V_{24}=360/24=15\,\text{V}\); \(45+30+15=90\,\text{V}\), which equals the battery voltage.)
Answer Details
(a) To say that the capacitance of a capacitor is \(5\,\mu\text{F}\) means that the capacitor stores a charge of \(5\,\mu\text{C}\) \((5\times10^{-6}\,\text{C})\) on each plate for every \(1\,\text{V}\) of potential difference applied across its plates. In other words, the ratio of the charge stored to the potential difference across the plates is \(5\,\mu\text{C V}^{-1}\), since \(C=\dfrac{Q}{V}\).
(b)(i) & (ii) The capacitance of a parallel-plate capacitor is given by \(C=\dfrac{\varepsilon A}{d}\). The factors on which it depends and the form of dependence are:
| Factor | How the capacitance depends on it |
| Common (overlapping) area of the plates, \(A\) | Directly proportional: \(C\propto A\). Increasing the area increases the capacitance. |
| Distance between the plates, \(d\) | Inversely proportional: \(C\propto \dfrac{1}{d}\). Increasing the separation decreases the capacitance. |
| Permittivity (nature) of the dielectric, \(\varepsilon\) | Directly proportional: \(C\propto \varepsilon\). A dielectric of higher permittivity increases the capacitance. |
(c)(i) Open-circuit diagram (switch open) showing the three capacitors in series with the \(90\,\text{V}\) battery:
(c)(ii) Effective capacitance. For capacitors in series the reciprocals add:
\[\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{3+2+1}{24}=\frac{6}{24}=\frac{1}{4}\]\[C=4\,\mu\text{F}\](c)(iii) Charge on each capacitor. In a series arrangement the same charge flows onto every capacitor, and it equals the charge on the equivalent capacitor:
\[Q=CV=4\times10^{-6}\times90=3.6\times10^{-4}\,\text{C}=360\,\mu\text{C}\]Therefore the charge on each of the \(8\,\mu\text{F}\), \(12\,\mu\text{F}\) and \(24\,\mu\text{F}\) capacitors is the same: \(Q=360\,\mu\text{C}\).
(c)(iv) P.d. across the \(8\,\mu\text{F}\) capacitor.
\[V_{8}=\frac{Q}{C_1}=\frac{360\,\mu\text{C}}{8\,\mu\text{F}}=45\,\text{V}\]The potential difference across the \(8\,\mu\text{F}\) capacitor is \(45\,\text{V}\). (As a check: \(V_{12}=360/12=30\,\text{V}\) and \(V_{24}=360/24=15\,\text{V}\); \(45+30+15=90\,\text{V}\), which equals the battery voltage.)
Question 7 Report
(a) Explain how a gas can be made to conduct electricity.
(b) Name the electric charge carriers in gases.
(a) How a gas can be made to conduct electricity
At ordinary pressure a gas is an insulator. It can be made to conduct by ionising it, that is by producing charged particles within it. This is done by applying a very high potential difference across the gas while it is at low pressure (as in a discharge tube), or by heating it strongly, or by exposing it to X-rays, ultraviolet or radioactive radiation. The high field/energy strips electrons from some of the gas atoms, producing free electrons and positive ions; under the applied field these charges drift to the electrodes and so a current flows.
(b) Electric charge carriers in gases
- Positive ions.
- Electrons (together with negative ions).
Answer Details
(a) How a gas can be made to conduct electricity
At ordinary pressure a gas is an insulator. It can be made to conduct by ionising it, that is by producing charged particles within it. This is done by applying a very high potential difference across the gas while it is at low pressure (as in a discharge tube), or by heating it strongly, or by exposing it to X-rays, ultraviolet or radioactive radiation. The high field/energy strips electrons from some of the gas atoms, producing free electrons and positive ions; under the applied field these charges drift to the electrodes and so a current flows.
(b) Electric charge carriers in gases
- Positive ions.
- Electrons (together with negative ions).
Question 8 Report
(a) What is electrolysis?
(b) A current of 2A is passed through a copper voltameter for 5 minutes. If the electrochemical equivalent of copper is 3.27 x 10\(^{-7}\) kg. Determine the mass of the corper deposited.
(a) What is electrolysis?
Electrolysis is the chemical decomposition of an electrolyte (a molten ionic compound or an aqueous solution of one) caused by the passage of a direct electric current through it. The current makes the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and new substances are set free.
(b) Mass of copper deposited
This uses Faraday's first law of electrolysis, which states that the mass \(m\) of a substance liberated at an electrode is proportional to the quantity of charge \(Q\) passed, so that \(m = zQ = zIt\), where \(z\) is the electrochemical equivalent.
Given: current \(I = 2\ \text{A}\); time \(t = 5\ \text{min} = 5 \times 60 = 300\ \text{s}\); electrochemical equivalent \(z = 3.27 \times 10^{-7}\ \text{kg C}^{-1}\).
Working:
\[ m = zIt = (3.27 \times 10^{-7}) \times 2 \times 300 \] \[ m = (3.27 \times 10^{-7}) \times 600 \] \[ m = 1.962 \times 10^{-4}\ \text{kg} \]The mass of copper deposited is \(1.962 \times 10^{-4}\ \text{kg}\), which is about \(0.196\ \text{g}\).
Key point: the time must be converted from minutes to seconds before substituting, because the ampere is one coulomb per second; forgetting to change 5 min into 300 s is the most common error and would give an answer 60 times too small.
Answer Details
(a) What is electrolysis?
Electrolysis is the chemical decomposition of an electrolyte (a molten ionic compound or an aqueous solution of one) caused by the passage of a direct electric current through it. The current makes the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and new substances are set free.
(b) Mass of copper deposited
This uses Faraday's first law of electrolysis, which states that the mass \(m\) of a substance liberated at an electrode is proportional to the quantity of charge \(Q\) passed, so that \(m = zQ = zIt\), where \(z\) is the electrochemical equivalent.
Given: current \(I = 2\ \text{A}\); time \(t = 5\ \text{min} = 5 \times 60 = 300\ \text{s}\); electrochemical equivalent \(z = 3.27 \times 10^{-7}\ \text{kg C}^{-1}\).
Working:
\[ m = zIt = (3.27 \times 10^{-7}) \times 2 \times 300 \] \[ m = (3.27 \times 10^{-7}) \times 600 \] \[ m = 1.962 \times 10^{-4}\ \text{kg} \]The mass of copper deposited is \(1.962 \times 10^{-4}\ \text{kg}\), which is about \(0.196\ \text{g}\).
Key point: the time must be converted from minutes to seconds before substituting, because the ampere is one coulomb per second; forgetting to change 5 min into 300 s is the most common error and would give an answer 60 times too small.
Question 9 Report
(a) State two properties of cathode rays.
(b) Explain how the intensity and energy of cathode rays may be increased.
(a) Two properties of cathode rays
- They travel in straight lines from the cathode and cast sharp shadows of objects placed in their path.
- They are streams of negatively charged particles (electrons), so they are deflected by both electric and magnetic fields (towards the positive plate).
Other acceptable properties: they carry energy and momentum, they produce heat and fluorescence on striking a screen, and they can produce X-rays on hitting a metal target.
(b) Increasing the intensity and energy of cathode rays
Intensity (number of electrons per second) is increased by raising the filament (heater) current. A hotter cathode emits more electrons by thermionic emission, so the beam current and hence the brightness increase.
Energy (speed of each electron) is increased by raising the accelerating potential difference between the cathode and the anode. Each electron gains kinetic energy \(eV\) as it is accelerated, so a larger anode voltage gives faster, more energetic electrons.
Answer Details
(a) Two properties of cathode rays
- They travel in straight lines from the cathode and cast sharp shadows of objects placed in their path.
- They are streams of negatively charged particles (electrons), so they are deflected by both electric and magnetic fields (towards the positive plate).
Other acceptable properties: they carry energy and momentum, they produce heat and fluorescence on striking a screen, and they can produce X-rays on hitting a metal target.
(b) Increasing the intensity and energy of cathode rays
Intensity (number of electrons per second) is increased by raising the filament (heater) current. A hotter cathode emits more electrons by thermionic emission, so the beam current and hence the brightness increase.
Energy (speed of each electron) is increased by raising the accelerating potential difference between the cathode and the anode. Each electron gains kinetic energy \(eV\) as it is accelerated, so a larger anode voltage gives faster, more energetic electrons.
Question 10 Report
(a) The mass and wavelength of a moving electron are 9.0 x 10\(^{-31}\) kg and 1.0 x 10\(^{-10}\)m respectively. Calculate the kinetic energy of the electron. [ h = 6.6 x 10\(^{-34}\) Js]
Kinetic energy of the electron from its de Broglie wavelength
The de Broglie relation links wavelength to momentum:
\[ \lambda = \frac{h}{mv} \quad\Rightarrow\quad v = \frac{h}{m\lambda} \]Substituting \(h = 6.6\times10^{-34}\,\text{Js}\), \(m = 9.0\times10^{-31}\,\text{kg}\), \(\lambda = 1.0\times10^{-10}\,\text{m}\):
\[ v = \frac{6.6\times10^{-34}}{(9.0\times10^{-31})(1.0\times10^{-10})} = \frac{6.6\times10^{-34}}{9.0\times10^{-41}} = 7.33\times10^{6}\,\text{ms}^{-1} \]Then the kinetic energy is:
\[ E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(9.0\times10^{-31})(7.33\times10^{6})^2 \] \[ E_k = \tfrac{1}{2}(9.0\times10^{-31})(5.38\times10^{13}) \approx 2.42\times10^{-17}\,\text{J} \]Equivalently, using \(E_k = \dfrac{h^2}{2m\lambda^2}\) gives the same result, \(E_k \approx 2.4\times10^{-17}\,\text{J}\).
Answer Details
Kinetic energy of the electron from its de Broglie wavelength
The de Broglie relation links wavelength to momentum:
\[ \lambda = \frac{h}{mv} \quad\Rightarrow\quad v = \frac{h}{m\lambda} \]Substituting \(h = 6.6\times10^{-34}\,\text{Js}\), \(m = 9.0\times10^{-31}\,\text{kg}\), \(\lambda = 1.0\times10^{-10}\,\text{m}\):
\[ v = \frac{6.6\times10^{-34}}{(9.0\times10^{-31})(1.0\times10^{-10})} = \frac{6.6\times10^{-34}}{9.0\times10^{-41}} = 7.33\times10^{6}\,\text{ms}^{-1} \]Then the kinetic energy is:
\[ E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(9.0\times10^{-31})(7.33\times10^{6})^2 \] \[ E_k = \tfrac{1}{2}(9.0\times10^{-31})(5.38\times10^{13}) \approx 2.42\times10^{-17}\,\text{J} \]Equivalently, using \(E_k = \dfrac{h^2}{2m\lambda^2}\) gives the same result, \(E_k \approx 2.4\times10^{-17}\,\text{J}\).
Question 11 Report
(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are produced. Represent this transmutation in a symbolic equation.
(b)(i) How does a radioactive atom differ from a stable one?
(ii) Explain 'half life'.
(iii) A sample of radioactive material has a haft life of 35 days. Calculate the fraction of the original quantity that will remain after 105 days.
(c) Light of wavelength 5.00 x 10\(^{-7}\)m is incident on a material of work function 1.90 eV. Calculate
(i) photon energy.
(ii) kinetic energy of the most energetic photo electron.
(iii) stopping potential [Plancks constant h =6.6 x 10\(^{-34}\)Js] [c= 3.0 x10\(^8\)ms\(^{-2}\), leV= 1.6 x 10\(^{19}\)J]
(a) Transmutation equation
Nitrogen-14 bombarded by a neutron emits an alpha particle:
\[ {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n} \;\rightarrow\; {}^{4}_{2}\text{He} + {}^{11}_{5}\text{B} \](Mass numbers: \(14+1 = 4+11\); atomic numbers: \(7+0 = 2+5\). The residual nucleus is boron-11.)
(b)(i) A radioactive atom has an unstable nucleus that spontaneously disintegrates, emitting radiation (alpha, beta or gamma), whereas a stable atom has a nucleus that does not decay.
(b)(ii) Half life is the time taken for half the atoms (or nuclei) in a given sample of a radioactive material to decay.
(b)(iii) \(105\,\text{days} = 3\times 35\,\text{days} = 3\) half lives. Fraction remaining:
\[ \left(\tfrac{1}{2}\right)^3 = \frac{1}{8} \](c) Photoelectric calculations (\(\lambda = 5.00\times10^{-7}\,\text{m}\), \(W_0 = 1.90\,\text{eV}\)):
(i) Photon energy:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{5.0\times10^{-7}} = 3.96\times10^{-19}\,\text{J} \ (\approx 2.48\,\text{eV}) \](ii) Maximum kinetic energy: work function \(W_0 = 1.90\times1.6\times10^{-19} = 3.04\times10^{-19}\,\text{J}\).
\[ E_k = E - W_0 = 3.96\times10^{-19} - 3.04\times10^{-19} = 9.2\times10^{-20}\,\text{J} \ (0.575\,\text{eV}) \](iii) Stopping potential:
\[ V_s = \frac{E_k}{e} = \frac{9.2\times10^{-20}}{1.6\times10^{-19}} = 0.575\,\text{V} \]Answer Details
(a) Transmutation equation
Nitrogen-14 bombarded by a neutron emits an alpha particle:
\[ {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n} \;\rightarrow\; {}^{4}_{2}\text{He} + {}^{11}_{5}\text{B} \](Mass numbers: \(14+1 = 4+11\); atomic numbers: \(7+0 = 2+5\). The residual nucleus is boron-11.)
(b)(i) A radioactive atom has an unstable nucleus that spontaneously disintegrates, emitting radiation (alpha, beta or gamma), whereas a stable atom has a nucleus that does not decay.
(b)(ii) Half life is the time taken for half the atoms (or nuclei) in a given sample of a radioactive material to decay.
(b)(iii) \(105\,\text{days} = 3\times 35\,\text{days} = 3\) half lives. Fraction remaining:
\[ \left(\tfrac{1}{2}\right)^3 = \frac{1}{8} \](c) Photoelectric calculations (\(\lambda = 5.00\times10^{-7}\,\text{m}\), \(W_0 = 1.90\,\text{eV}\)):
(i) Photon energy:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{5.0\times10^{-7}} = 3.96\times10^{-19}\,\text{J} \ (\approx 2.48\,\text{eV}) \](ii) Maximum kinetic energy: work function \(W_0 = 1.90\times1.6\times10^{-19} = 3.04\times10^{-19}\,\text{J}\).
\[ E_k = E - W_0 = 3.96\times10^{-19} - 3.04\times10^{-19} = 9.2\times10^{-20}\,\text{J} \ (0.575\,\text{eV}) \](iii) Stopping potential:
\[ V_s = \frac{E_k}{e} = \frac{9.2\times10^{-20}}{1.6\times10^{-19}} = 0.575\,\text{V} \]Question 12 Report
(a) In his first attempt, a long jumper took off from the springboard with a speed of 8 ms\(^{-1}\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \(\frac{v^2 sin \theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt.
(b)(i) State Hooke's law of elasticity.
(ii) Describe an experiment to verify Hooke's law.
(iii) State two precautions you would take if you were to perform this experiment in the laboratory.
(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\(^{-2}\)]
(a) Gain in the long jump
The horizontal range of a projectile is \(R = \dfrac{v^2\sin 2\theta}{g}\). With \(v = 8\,\text{ms}^{-1}\) and \(g = 10\,\text{ms}^{-2}\), \(\dfrac{v^2}{g} = \dfrac{64}{10} = 6.4\,\text{m}\).
First attempt (\(\theta = 30^\circ\)): \(R_1 = 6.4\sin 60^\circ = 6.4(0.8660) = 5.5424\,\text{m}\).
Second attempt (\(\theta = 45^\circ\)): \(R_2 = 6.4\sin 90^\circ = 6.4(1) = 6.4000\,\text{m}\).
\[ R_2 - R_1 = 6.4000 - 5.5424 = 0.8576\,\text{m} \quad\text{(shown)} \](b)(i) Hooke's law: Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load).
(b)(ii) Experiment to verify Hooke's law: A spiral spring is clamped vertically with a pointer against a metre rule. The unstretched pointer reading is noted. Known masses are added one at a time; for each load the new pointer reading is taken and the extension found by subtraction. A graph of load against extension is plotted; a straight line through the origin verifies that extension is proportional to load, confirming Hooke's law.
(b)(iii) Two precautions:
- Do not exceed the elastic limit of the spring (avoid overloading so it returns to its original length).
- Take readings with the eye level to the pointer to avoid parallax error; allow the spring to settle before reading.
(c) Mass of the scale pan
Extension is proportional to the total load (pan mass \(m\) plus object). Natural length \(= 20.00\,\text{cm}\).
With 40 g: length \(21.80\,\text{cm}\), extension \(e_1 = 1.80\,\text{cm}\), load \(=(m+40)\).
With 60 g: length \(22.05\,\text{cm}\), extension \(e_2 = 2.05\,\text{cm}\), load \(=(m+60)\).
\[ \frac{m+40}{1.80} = \frac{m+60}{2.05} \] \[ 2.05(m+40) = 1.80(m+60) \] \[ 2.05m + 82 = 1.80m + 108 \;\Rightarrow\; 0.25m = 26 \;\Rightarrow\; m = 104\,\text{g} \]The mass of the scale pan is \(104\,\text{g}\ (0.104\,\text{kg})\).
Answer Details
(a) Gain in the long jump
The horizontal range of a projectile is \(R = \dfrac{v^2\sin 2\theta}{g}\). With \(v = 8\,\text{ms}^{-1}\) and \(g = 10\,\text{ms}^{-2}\), \(\dfrac{v^2}{g} = \dfrac{64}{10} = 6.4\,\text{m}\).
First attempt (\(\theta = 30^\circ\)): \(R_1 = 6.4\sin 60^\circ = 6.4(0.8660) = 5.5424\,\text{m}\).
Second attempt (\(\theta = 45^\circ\)): \(R_2 = 6.4\sin 90^\circ = 6.4(1) = 6.4000\,\text{m}\).
\[ R_2 - R_1 = 6.4000 - 5.5424 = 0.8576\,\text{m} \quad\text{(shown)} \](b)(i) Hooke's law: Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load).
(b)(ii) Experiment to verify Hooke's law: A spiral spring is clamped vertically with a pointer against a metre rule. The unstretched pointer reading is noted. Known masses are added one at a time; for each load the new pointer reading is taken and the extension found by subtraction. A graph of load against extension is plotted; a straight line through the origin verifies that extension is proportional to load, confirming Hooke's law.
(b)(iii) Two precautions:
- Do not exceed the elastic limit of the spring (avoid overloading so it returns to its original length).
- Take readings with the eye level to the pointer to avoid parallax error; allow the spring to settle before reading.
(c) Mass of the scale pan
Extension is proportional to the total load (pan mass \(m\) plus object). Natural length \(= 20.00\,\text{cm}\).
With 40 g: length \(21.80\,\text{cm}\), extension \(e_1 = 1.80\,\text{cm}\), load \(=(m+40)\).
With 60 g: length \(22.05\,\text{cm}\), extension \(e_2 = 2.05\,\text{cm}\), load \(=(m+60)\).
\[ \frac{m+40}{1.80} = \frac{m+60}{2.05} \] \[ 2.05(m+40) = 1.80(m+60) \] \[ 2.05m + 82 = 1.80m + 108 \;\Rightarrow\; 0.25m = 26 \;\Rightarrow\; m = 104\,\text{g} \]The mass of the scale pan is \(104\,\text{g}\ (0.104\,\text{kg})\).
Question 13 Report
(a)(i) What is an echo? (ii) State two useful applications of echoes.
(iii) Why are the walls, floors and ceilings of a recording studio heavily padded?
(b)(i) Explain timbre and overtones.
(ii) What is resonance?
(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\(^{-1}\), calculate the velocity at which the ship was approaching the cliff.
(a)(i) An echo is a reflected sound wave heard distinctly after the original sound, produced when sound is reflected from a hard, distant surface back to the listener.
(a)(ii) Two useful applications:
- Depth sounding (echo sounding/SONAR) to measure the depth of the sea or locate shoals of fish and submarines.
- Detecting flaws in metals and, in medicine, ultrasonic scanning of the body.
(a)(iii) The walls, floors and ceilings of a recording studio are heavily padded with soft, porous material to absorb sound and prevent reflection (echoes and reverberation), so that only clear, direct sound is recorded.
(b)(i) Timbre (quality) is the property of a musical note that lets the ear distinguish two notes of the same pitch and loudness from different sources; it depends on the number and relative strength of the overtones present. Overtones are the higher frequencies (above the fundamental) that accompany a note.
(b)(ii) Resonance is the condition in which a body is set into vibration with large amplitude by a periodic force whose frequency equals the natural frequency of the body.
(c) Velocity of the ship
Distance of the cliff at the first sounding \(= \tfrac{1}{2}(340)(12) = 2040\,\text{m}\).
Distance at the second sounding \(= \tfrac{1}{2}(340)(8) = 1360\,\text{m}\).
Distance travelled by the ship between the two soundings:
\[ 2040 - 1360 = 680\,\text{m} \]Time between soundings \(= 2.1\,\text{min} = 126\,\text{s}\). Hence:
\[ v = \frac{680}{126} \approx 5.4\,\text{ms}^{-1} \]The ship approaches the cliff at about \(5.4\,\text{ms}^{-1}\).
Answer Details
(a)(i) An echo is a reflected sound wave heard distinctly after the original sound, produced when sound is reflected from a hard, distant surface back to the listener.
(a)(ii) Two useful applications:
- Depth sounding (echo sounding/SONAR) to measure the depth of the sea or locate shoals of fish and submarines.
- Detecting flaws in metals and, in medicine, ultrasonic scanning of the body.
(a)(iii) The walls, floors and ceilings of a recording studio are heavily padded with soft, porous material to absorb sound and prevent reflection (echoes and reverberation), so that only clear, direct sound is recorded.
(b)(i) Timbre (quality) is the property of a musical note that lets the ear distinguish two notes of the same pitch and loudness from different sources; it depends on the number and relative strength of the overtones present. Overtones are the higher frequencies (above the fundamental) that accompany a note.
(b)(ii) Resonance is the condition in which a body is set into vibration with large amplitude by a periodic force whose frequency equals the natural frequency of the body.
(c) Velocity of the ship
Distance of the cliff at the first sounding \(= \tfrac{1}{2}(340)(12) = 2040\,\text{m}\).
Distance at the second sounding \(= \tfrac{1}{2}(340)(8) = 1360\,\text{m}\).
Distance travelled by the ship between the two soundings:
\[ 2040 - 1360 = 680\,\text{m} \]Time between soundings \(= 2.1\,\text{min} = 126\,\text{s}\). Hence:
\[ v = \frac{680}{126} \approx 5.4\,\text{ms}^{-1} \]The ship approaches the cliff at about \(5.4\,\text{ms}^{-1}\).
Question 14 Report
(a) What is meant by 'a beam of polarised light?
(b) With the aid of well labelled diagrams, illustrate the action in of a polaroid spectacle on a beam of sunlight.
(a) A beam of polarised light
Light is a transverse wave. In an ordinary (unpolarised) beam the electric-field vibrations occur in all planes perpendicular to the direction in which the light travels. A beam of polarised light is a beam in which these vibrations have been restricted to one single plane perpendicular to the direction of travel. In other words, the vibration of the wave takes place in only one plane (the plane of polarisation), so the light is said to be plane-polarised.
(b) Action of a polaroid spectacle on a beam of sunlight
Sunlight reaching the eye is unpolarised: its vibrations lie in every plane at right angles to the beam. A polaroid lens contains long parallel molecular chains that create a single transmission (polarising) axis. When sunlight strikes the polaroid:
- only the vibration parallel to the transmission axis is allowed through, and
- the vibrations in every other plane are absorbed.
The light that emerges is therefore plane-polarised and much reduced in intensity. This is why polaroid spectacles cut down the dazzling glare of sunlight (especially reflected glare, which is already partly polarised), so the wearer sees more comfortably. The action is illustrated below.
In the diagram, the incoming sunlight vibrates in all directions (shown by the double-headed arrows pointing every way). After passing through the polaroid, whose transmission axis is vertical, only the vertical component of vibration survives, so the beam reaching the eye is plane-polarised in the vertical plane and dimmer than the original sunlight.
Answer Details
(a) A beam of polarised light
Light is a transverse wave. In an ordinary (unpolarised) beam the electric-field vibrations occur in all planes perpendicular to the direction in which the light travels. A beam of polarised light is a beam in which these vibrations have been restricted to one single plane perpendicular to the direction of travel. In other words, the vibration of the wave takes place in only one plane (the plane of polarisation), so the light is said to be plane-polarised.
(b) Action of a polaroid spectacle on a beam of sunlight
Sunlight reaching the eye is unpolarised: its vibrations lie in every plane at right angles to the beam. A polaroid lens contains long parallel molecular chains that create a single transmission (polarising) axis. When sunlight strikes the polaroid:
- only the vibration parallel to the transmission axis is allowed through, and
- the vibrations in every other plane are absorbed.
The light that emerges is therefore plane-polarised and much reduced in intensity. This is why polaroid spectacles cut down the dazzling glare of sunlight (especially reflected glare, which is already partly polarised), so the wearer sees more comfortably. The action is illustrated below.
In the diagram, the incoming sunlight vibrates in all directions (shown by the double-headed arrows pointing every way). After passing through the polaroid, whose transmission axis is vertical, only the vertical component of vibration survives, so the beam reaching the eye is plane-polarised in the vertical plane and dimmer than the original sunlight.
Question 15 Report
The value of the e.m.f of a voltaic cell, which has dilute tetraoxosulphate (VI) acid as its electrolyte and copper and zinc as its electrodes becomes less with use. Explain this observation and state how it can be corrected.
The fall in e.m.f. with use is caused by two defects of the simple voltaic cell:
- Polarisation. As the cell works, hydrogen gas is liberated and clings to the copper (positive) plate, forming an insulating layer. This layer increases the internal resistance and sets up a back e.m.f., so the useful e.m.f. falls.
- Local action. Impurities (such as carbon and iron) in the commercial zinc plate form tiny local cells with the zinc, so the zinc is eaten away even when no current is being drawn, weakening the cell.
Corrections:
- Polarisation is removed by adding a depolariser (an oxidising agent such as manganese(IV) oxide or potassium dichromate) which oxidises the hydrogen to water.
- Local action is prevented by amalgamating the zinc plate, i.e. coating it with mercury so that the surface presents a clean layer of pure zinc only.
Answer Details
The fall in e.m.f. with use is caused by two defects of the simple voltaic cell:
- Polarisation. As the cell works, hydrogen gas is liberated and clings to the copper (positive) plate, forming an insulating layer. This layer increases the internal resistance and sets up a back e.m.f., so the useful e.m.f. falls.
- Local action. Impurities (such as carbon and iron) in the commercial zinc plate form tiny local cells with the zinc, so the zinc is eaten away even when no current is being drawn, weakening the cell.
Corrections:
- Polarisation is removed by adding a depolariser (an oxidising agent such as manganese(IV) oxide or potassium dichromate) which oxidises the hydrogen to water.
- Local action is prevented by amalgamating the zinc plate, i.e. coating it with mercury so that the surface presents a clean layer of pure zinc only.
